Timeline for Define $\mathbb{N}$ in the ring $\mathbb{Z}$ without Lagrange's theorem
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| S Nov 10, 2016 at 14:16 | history | suggested | Loreno Heer | CC BY-SA 3.0 |
added missing in to the title
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| Nov 10, 2016 at 13:56 | review | Suggested edits | |||
| S Nov 10, 2016 at 14:16 | |||||
| Sep 10, 2016 at 14:43 | comment | added | Sergei Akbarov | @LSpice, excuse me, I did not understand from the very beginning, what the question is about, and besides this I forgot to write another condition: $1\in X$. | |
| Sep 10, 2016 at 13:24 | comment | added | LSpice | @SergeiAkbarov, probably I misunderstand, but minimal in what sense? If in the sense of set containment, then replacing $\mathbb N$ by its subset of successors seems to contradict its minimality. | |
| Sep 10, 2016 at 13:14 | vote | accept | CommunityBot | ||
| Sep 10, 2016 at 12:53 | answer | added | Sidney Raffer | timeline score: 35 | |
| Sep 10, 2016 at 9:56 | answer | added | Wojowu | timeline score: 23 | |
| Sep 10, 2016 at 9:52 | comment | added | Sergei Akbarov | Ah, excuse me, I did not understand the question! | |
| Sep 10, 2016 at 9:50 | comment | added | Wojowu | @SergeiAkbarov This is not a first-order definition. | |
| Sep 10, 2016 at 9:50 | comment | added | Sergei Akbarov | Why don't you want to define $\mathbb N$ as a minimal subset $X$ in $\mathbb Z$ with the property $x\in X\ \Rightarrow\ x+1\in X$? | |
| Sep 10, 2016 at 9:46 | comment | added | KConrad | Is this just asked out of curiosity, or do you have a problem with this use of the four-square theorem? I always regarded this approach as illustrating that the result (definability of $\mathbf N$ within $\mathbf Z$) has real substance to it. | |
| Sep 10, 2016 at 9:09 | history | asked | user40023 | CC BY-SA 3.0 |